the problem$\displaystyle

\int 2t/(t-3)^2$

solving with partial fractions$\displaystyle

\frac{a}{t-3} + \frac{a}{(t-3)^2}$

$\displaystyle t = 3, b=6$

$\displaystyle t=0, a =2$

$\displaystyle

\int \frac{2}{t-3} + \int \frac{6}{(t-3)^2}$

$\displaystyle = 2lnt-3 + \int \frac{6}{(t-3)^2}$

but I don't know how to solve for $\displaystyle \int \frac{6}{(t-3)^2}$